PROBLEM 1: NON-NEWTONIAN FLUID MECHANICS
A very large class of fluids that we encounter in our lives are non-Newtonian. Examples include
shampoo, toothpaste, ketchup, mayonnaise, bio]macromolecule (e.g. DNA, collagen, polymers)
solutions, nanopartiele suspensions, creams, foams,
and various gels. For these fluids, the simple
relationship between the shear rate and the shear
stress does not hold. Instead, additional, more
involved physics need to be taken into account to
capture the fluid's
by advection (flow)
behavior. In this problem, we want to explore the
implications of this additional physics for a simple
case of laminar flow on a vertical wall.
Consider a liquid film that is flowing down a
by molecular transport
vertical surface in a fully developed profile. The
thickness of the fluid film is 8 = 2 mm. A
differential momentum balance for a fluid element
(see Figure) yields the following differential
equation for the shear stress:
by advection (flow)
where Taz is the shear stress, or the flux of momentum in the x-direction as a result of flow in the
z-direction, is the fluid density, and g is the gravitational acceleration. The relevant boundary
condition here is that at the liquid/air interface (x = o), the shear stress is zero (this is because
air, which has extremely low viscosity compared to most liquids and complex fluids, provides
almost no viscous resistance to flow of the liquid down the wall).
This conservation law (momentum balance) must be solved in conjunction with a constitutive
equation. In Fluid Mechanics, the constitutive equation must relate the flux of momentum to
fluid properties. For a Newtonian fluid (such as water, honey, ethanol, and some low molecular
weight polymers), this constitutive equation is the familiar Newton's law of viscosity:
where u is the fluid viscosity (constant for a Newtonian fluid), and V2 is the local fluid velocity in
the z-direction. The negative sign is needed to account for the direction of transport. For a class
of non-Newtonian fluids called Bingham plastics (or Bingham fluids - ketchup and mayonnaise
are examples), the constitutive equation is:
if t. >to
where To is a yield stress and Ho is an effective viscosity. This is like the behavior that you expect
for ketchup, where it won't flow out of the bottle with just a gentle tilt, but if you give it a bump
while tilted, all of a sudden it can flow out almost like a Newtonian fluid.
Finally, for yet another class of fluids called power-law fluids, the constitutive equation is:
where K is a parameter with units of N.st/m² and the exponent n is called the flow index. For a
pseudoplastic (shear-thinning) fluid such as solutions of macromolecules (e.g. xanthan gum,
polymers, etc) in another solvent, we have n < 1, and for a dilatant (shear-thickening) fluid such
as Silly Putty, concentrated nanoparticle suspensions, or mud, we have n> 1.
The relevant boundary condition in this problem for any of these constitutive equations is that
the fluid layer right adjacent to the wall will stick to the wall (this is called a no-slip boundary
condition). Therefore, at the liquid/wall interface (x - 8), the fluid velocity is zero.
Finally, the fluid's volumetric flow rate flowing down the wall can be calculated from:
a. On two separate graphs, plot the velocity profile and shear stress (both as functions of x) for a
Newtonian fluid with 8.937 x 10-4 kg/m.s and p= 998 kg/m³ (like water).
b. Calculate the volumetric flow rate for part a.
c. On two separate graphs, plot the velocity profile and shear stress (both as functions of x) for a
Newtonian fluid with = 10.211 kg/m.s and P = 1300 kg/m³ (like honey).
d. Calculate the volumetric flow rate for part c.
e. On two separate graphs, plot the velocity profile and shear stress (both as functions of x) for a
pseudoplastic fluid with K - 0.01 N.st/m², n - 0.5, and P 1123 kg/m³.
f. Calculate the volumetric flow rate for part e.
g. On two separate graphs, plot the velocity profile and shear stress (both as functions of x) for a
Bingham fluid with to 6.0 kg/m.s², o= 0.2 kg/m.s and p 803 kg/m³.
h. Calculate the volumetric flow rate for part g.
i. Which fluids have the highest and lowest volumetric flow rates? Can you guess why?
j. What is unique about the velocity profile for the Bingham fluid?
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clear; clc; close all;
tau_0 = 0;%zero shear boundary condition
v_0 = 0;%Boundary layer condition
X0 = [v_0 tau_0];
global rho g mu tau0 mu0 n K
g = 9.81;
delta = 2e-3;%Thickness of layer
xspan = linspace(0,delta,100);
mu1 = 8.937E-4;
rho1 = 998;
mu2 = 10.211;
rho2 = 1300;
K3 = 0.01;
n3 = 0.5;
rho3 = 1123;
tau04 = 6;
mu04 = 0.2;
rho4 = 803;
rho = rho1; mu = mu1;
fluidType = 'Newtonian';
[s1_x,s1_X] = ode45(@(t,X) fmshearsolve(t,X,fluidType), xspan, X0);
stra = 'Velocity Profile';
strb = 'Shear stress profile';
s1_str = 'Case 1(water like):';
str11 = [s1_str,' ',fluidType,' ',stra];
str12 = [s1_str,' ',fluidType,' ',strb];
plot(s1_x*1000,s1_X(:,1),'LineWidth',2,'Color',[1 0 0]);
xlabel('x(mm)'); ylabel('Velocity(m/s)'); title(str11);
plot(s1_x*1000,s1_X(:,2),'LineWidth',2,'Color',[1 0 0]);
xlabel('x(mm)'); ylabel('Shear stress(kg/m.s^2)'); title(str12);
s1_Vdot = trapz(s1_x,s1_X(:,1));
fprintf('\n\nThe volumetric flow rate for Case 1(water like) is computed as %10.8fm^3/s for unit depth.\n\n',s1_Vdot);...